How do you solve the inequality #abs(2-3x)>=2/3#?

1 Answer
Jul 26, 2017

Answer:

See a solution process below: #(-oo, 4/9]#; #[8/9, +oo)#

Explanation:

The absolute value function takes any negative or positive term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

#-2/3 >= 2 - 3x >= 2/3#

First, subtract #color(red)(2)# from each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:

#-2/3 - color(red)(2) >= 2 - color(red)(2) - 3x >= 2/3 - color(red)(2)#

#-2/3 - (3/3 xx color(red)(2)) >= 0 - 3x >= 2/3 - (3/3 xx color(red)(2))#

#-2/3 - 6/3 >= -3x >= 2/3 - 6/3#

#-8/3 >= -3x >= -4/3#

Now, divide each segment by #color(blue)(-3)# to solve for #x# while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators:

#(-8/3)/color(blue)(-3) color(red)(<=) (-3x)/color(blue)(-3) color(red)(<=) (-4/3)/color(blue)(-3)#

#((-8)/3)/(color(blue)(-3)/1) color(red)(<=) (color(red)(cancel(color(black)(-3)))x)/cancel(color(blue)(-3)) color(red)(<=) ((-4)/3)/(color(blue)(-3)/1)#

#(-8 xx 1)/(3 xx color(blue)(-3)) color(red)(<=) x color(red)(<=) (-4 xx 1)/(3 xx color(blue)(-3))#

#(-8)/(-9) color(red)(<=) x color(red)(<=) (-4)/(-9)#

#8/9 color(red)(<=) x color(red)(<=) 4/9#

Or

#x <= 4/9#; #x >= 8/9#

Or, in interval notation:

#(-oo, 4/9]#; #[8/9, +oo)#